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Weierstrass Substitution Worked Examples cheat sheet - grade college

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Calculus Grade college

Weierstrass Substitution Worked Examples Cheat Sheet

A printable reference covering Weierstrass substitution, half-angle identities, rationalizing trigonometric integrals, and worked integral examples for college.

TopicsIntegralsTrigonometryAlgebraFunctionsCalculusStudy Guide
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The Weierstrass substitution rewrites trigonometric integrals using t=tan⁡x2t = \tan\frac{x}{2}. This changes many integrals involving sin⁡x\sin x and cos⁡x\cos x into rational functions of tt. A worked examples cheat sheet helps students recognize when the substitution is useful and apply the algebra without losing track of constants.

It is especially useful in college calculus when standard identities do not simplify an integral quickly.

The core formulas are sin⁡x=2t1+t2\sin x = \frac{2t}{1+t^2}, cos⁡x=1−t21+t2\cos x = \frac{1-t^2}{1+t^2}, and dx=21+t2 dtdx = \frac{2}{1+t^2}\,dt. After substitution, simplify the integrand into a rational expression and integrate using algebra, partial fractions, or basic antiderivatives. At the end, convert back using t=tan⁡x2t = \tan\frac{x}{2} unless the problem asks for an answer in terms of tt.

Key Facts

  • The Weierstrass substitution is t=tan⁡x2t = \tan\frac{x}{2}, which is also written as t=tan⁡(x2)t = \tan\left(\frac{x}{2}\right).
  • The sine conversion formula is sin⁡x=2t1+t2\sin x = \frac{2t}{1+t^2}.
  • The cosine conversion formula is cos⁡x=1−t21+t2\cos x = \frac{1-t^2}{1+t^2}.
  • The differential conversion formula is dx=21+t2 dtdx = \frac{2}{1+t^2}\,dt.
  • The tangent conversion formula is tan⁡x=2t1−t2\tan x = \frac{2t}{1-t^2} when 1−t2≠01-t^2 \neq 0.
  • An integral of the form ∫R(sin⁡x,cos⁡x) dx\int R(\sin x,\cos x)\,dx becomes ∫R(2t1+t2,1−t21+t2)21+t2 dt\int R\left(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2}\right)\frac{2}{1+t^2}\,dt.
  • For definite integrals, convert bounds using t=tan⁡x2t = \tan\frac{x}{2}, but watch for discontinuities when the interval crosses x=π+2πkx = \pi + 2\pi k.
  • After integrating in tt, substitute back with t=tan⁡x2t = \tan\frac{x}{2} to express the final answer in terms of xx.

Vocabulary

Weierstrass substitution
A trigonometric substitution using t=tan⁡x2t = \tan\frac{x}{2} to convert expressions involving sin⁡x\sin x and cos⁡x\cos x into rational functions.
Half-angle substitution
Another name for the substitution t=tan⁡x2t = \tan\frac{x}{2} because it uses the tangent of half the original angle.
Rational function
A function that can be written as a quotient of polynomials, such as t2+1t−3\frac{t^2+1}{t-3}.
Back-substitution
The step of replacing tt with tan⁡x2\tan\frac{x}{2} after finding an antiderivative in terms of tt.
Partial fractions
A method for rewriting a rational expression as simpler fractions that are easier to integrate.
Domain restriction
A limitation on where a substitution is valid, such as avoiding points where t=tan⁡x2t = \tan\frac{x}{2} is undefined.

Common Mistakes to Avoid

  • Forgetting to replace dxdx is wrong because the substitution is incomplete without dx=21+t2 dtdx = \frac{2}{1+t^2}\,dt.
  • Using sin⁡x=t1+t2\sin x = \frac{t}{1+t^2} is wrong because the correct numerator is 2t2t, so the correct formula is sin⁡x=2t1+t2\sin x = \frac{2t}{1+t^2}.
  • Using cos⁡x=t2−11+t2\cos x = \frac{t^2-1}{1+t^2} reverses the sign; the correct formula is cos⁡x=1−t21+t2\cos x = \frac{1-t^2}{1+t^2}.
  • Changing definite bounds incorrectly gives the wrong area or accumulated value; each bound x=ax=a must become t=tan⁡a2t = \tan\frac{a}{2}.
  • Leaving the final indefinite integral in terms of tt can be incomplete when the original problem uses xx; substitute back with t=tan⁡x2t = \tan\frac{x}{2} unless instructed otherwise.

Practice Questions

  1. 1 Use t=tan⁡x2t = \tan\frac{x}{2} to evaluate ∫dx1+sin⁡x\int \frac{dx}{1+\sin x}.
  2. 2 Use the Weierstrass substitution to evaluate ∫dx2+cos⁡x\int \frac{dx}{2+\cos x}.
  3. 3 Convert ∫sin⁡x1+cos⁡x dx\int \frac{\sin x}{1+\cos x}\,dx into an integral in tt using t=tan⁡x2t = \tan\frac{x}{2}, then simplify before integrating.
  4. 4 Explain why the Weierstrass substitution is often useful for integrals involving rational combinations of sin⁡x\sin x and cos⁡x\cos x.

Understanding Weierstrass Substitution Worked Examples

The substitution has a geometric reason, not just a list of formulas to memorize. Picture the unit circle and the point at the far left, with coordinates negative one and zero. A line from that point meets the circle at the angle x.

Its slope can be used as a single parameter. That parameter is tangent of half the angle. Solving the circle equation with this slope produces rational expressions for the horizontal and vertical coordinates.

Since cosine and sine are those coordinates, trigonometry is replaced by ordinary fractions. This is why the method works so broadly. It is really a way of describing points on a circle using one real number.

In a worked problem, the important task is accounting for every factor. Students often replace sine and cosine correctly but forget that the differential contributes an extra fraction. Write the entire integrand after every replacement before trying to cancel anything.

Then combine fractions carefully. For example, an integral with one divided by two plus sine x becomes a fraction in t whose denominator can often be factored. At that stage it is no longer a trigonometry problem.

It may require division of polynomials, completing a square, or partial fractions. Choose the algebra method from the final rational expression, rather than choosing it from the original integral.

Partial fractions are common when the denominator splits into simple factors. A denominator involving one plus t squared usually leads to an inverse tangent term. A repeated linear factor leads to powers in the denominator, which integrate by the power rule.

Some answers contain logarithms because a factor such as t minus a constant appears in the denominator. These results can look unfamiliar after converting back, but they are valid antiderivatives. Constants can be absorbed into the final constant of integration.

It is worth differentiating the answer when possible. The chain rule during this check catches missing factors of two, incorrect signs, and errors made while simplifying.

Definite integrals need extra care because the parameter does not behave continuously through every angle. Tangent of half the angle becomes undefined at angles equivalent to pi. If an interval crosses such an angle, split the original integral at that point before changing variables.

Each piece then has bounds that map safely into t values. This issue matters in applications involving periodic motion, waves, rotating objects, and electrical signals, where an interval may cover a full cycle. In class, pay attention to the domain before doing algebra.

A clean rational integral does not guarantee that the substitution was valid across the whole original interval. Good solutions show the converted bounds, note any split interval, simplify systematically, then state the answer in the original variable.